Wait — max HEIGHT at 90°, not range formula at 45°
Max range at 45°: R=v₀²/g. Max height when θ=90° (vertical throw): H=v₀²/2g. So H_max = R_max/2 = 50 m. Answer: B, not C. Review carefully!
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Moderate Level — NEET / JEE Main
2–3 step problems. Conceptual traps included. Target: 7/10 correct. Time per question: 2–3 minutes.
JEE MainConceptual
Q5. From a tower of height 20 m, a ball is thrown horizontally. At the same instant, another ball is dropped from the same height. Which statement is correct?
A
The thrown ball reaches ground first
B
The dropped ball reaches ground first
C
Both reach ground simultaneously
D
Depends on the horizontal speed
1
Both start at same height, both have the same vertical initial velocity (zero), same g. Time = √(2h/g) = √(4) = 2 s for both. They land simultaneously. Horizontal motion has zero effect on vertical timing.
NEETSpeed at Point
Q6. A ball is thrown at 30° with 40 m/s. What is the speed of the ball at the highest point? (g = 10)
A
0
B
40 m/s
C
20√3 m/s
D
20 m/s
1
At highest point, vy = 0. Only horizontal velocity remains. vₓ = v₀cos30° = 40×(√3/2) = 20√3 m/s
JEE MainTime Period
Q7. A particle moves in a circle of radius R with constant speed v. The time period is T. If radius is doubled and speed is halved, new time period T' is:
Q8. A ball rolls off the edge of a table at 5 m/s. The table is 1.25 m high. How far from the base of the table does the ball land? (g = 10 m/s²)
A
1 m
B
1.5 m
C
2.5 m
D
3 m
1
t = √(2h/g) = √(2×1.25/10) = √(0.25) = 0.5 s
2
R = u×t = 5×0.5 = 2.5 m
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Advanced Level — JEE Advanced Style
Multi-correct, paragraph-based, integer-type. Think before calculating. One wrong step = wrong final answer. Target: 5/8 correct for top 1%.
JEE Adv Multi-CorrectProjectile Concepts
Q9 (Multi-correct): For a projectile launched at angle θ with speed v₀, which of the following is/are correct? [Select ALL that apply]
A
Horizontal velocity component remains constant throughout
B
Speed is minimum at the highest point when θ = 90°
C
The trajectory is a parabola when launched from ground level
D
Velocity is perpendicular to acceleration at the highest point
A ✓ — Correct. aₓ = 0 always. B ✗ — Wrong. When θ=90°, at highest point v=0 (truly minimum=0). But for any other angle, speed at highest point = v₀cosθ, which is minimum for that trajectory. B has poor wording — trick question. C ✓ — Correct. y = x tanθ − gx²/2v₀²cos²θ is parabola in x. D ✓ — Correct. At highest point, velocity is horizontal, acceleration (g) is vertical — perpendicular.
JEE AdvInteger Type
Q10 (Integer-type): A stone is thrown at 10√2 m/s at 45°. At a height of 5m, the horizontal distance from launch point is x m. Find the value of 2x. (g = 10 m/s²)
y = 5m: 5 = 10t − 5t² → t² − 2t + 1 = 0 → (t−1)² = 0 → t = 1 s
3
x = v₀cosθ × t = 10×1 = 10 m. 2x = 20
JEE AdvVertical Circle
Q11: A ball is tied to a string of length 2m. It is whirled in a vertical circle. What is the minimum speed at the top of the circle? If the speed at the top is 2√(gR), find the ratio of tension at bottom to tension at top. (g = 10 m/s²)
1
Min speed at top: v_t = √(gR) = √(10×2) = √20 = 2√5 m/s